牛顿法求平方根的逻辑有什么问题？

Question

我写了下面的代码，用牛顿法通过逐次逼近找到平方根，但它没有给我正确的答案。可以请某人解释它吗？

#include<stdio.h> 
#include<stdlib.h> 
#define square(x) x*x 
double rootByNewtonApprox(int n); 
double improve(double n); 
double average(double a,double b); 
int goodEnough(double guess); 
double guess(int n); 
int number; 
int main(void) 
{ 

    double root; 
    printf("\nEnter the number you want square root of: "); 
    scanf("%d",&number); 
    if(number<0) 
      number = -1* number; 
    root = rootByNewtonApprox(number); 
    printf("\nThe square root of %d is %lf\n",number,root); 
    return 0; 
} 
double guess(int n) 
{ 
    return n/2; 
} 
double rootByNewtonApprox(int n) 
{ 
    if(goodEnough(guess(n))) 
      return guess(n); 
    else 
      rootByNewtonApprox(improve(guess(n))); 
} 

double improve(double guess) 
{ 
    return average(guess,(number/guess)); 
} 
double average(double a,double b) 
{ 
    return ((a+b)/2); 
} 
int goodEnough(double guess) 
{ 
    if(abs(square(guess) - number) <= 0.001) 
      return 1; 
    else 
      return 0; 
} 

现在，当我给n = 2它给人的输出中nan，当我给n = 9它告诉segmentation Fault。

Answer 1

double guess(int n) 
{ 
    return n/(double) 2; 
} 

Answer 2

你忘了一个return

double rootByNewtonApprox(int n) 
{ 
if(goodEnough(guess(n))) 
    return guess(n); 
else 
    return rootByNewtonApprox(improve(guess(n))); 
    ^
}